Question: A particle moving in the $xy$ -plane has velocity vector given by $v(t)=\left(-5t^{4},-11t^{3}\right)$ for time $t\geq 0$. At $t=1$, the particle is at the point $(5,-5)$. What is the particle's position at $t=4$ ? $($
Explanation: To find the particle's position at $t=4$, we need to find its horizontal displacement $\Delta x$ and its vertical displacement $\Delta y$, and add those to its initial position $(5,-5)$ : $\text{Position at }t=4\text{: }(5+\Delta x,-5+\Delta y)$ The particle's horizontal displacement can be found by taking the definite integral of the horizontal component of $v(t)$ between $t=1$ and $t=4$ : $\Delta x=\int_{1}^{4} -5t^{4}\,dt=-1023$ The particle's vertical displacement can be found by taking the definite integral of the vertical component of $v(t)$ between $t=1$ and $t=4$ : $\Delta y=\int_{1}^{4} -11t^{3}\,dt=-\dfrac{2805}{4}$ Now we can find the particle's position: $\begin{aligned} &\phantom{=}(5+\Delta x,-5+\Delta y) \\\\ &=\left(5+(-1023),-5+\left(-\dfrac{2805}{4}\right)\right) \\\\ &=(-1018,-706.25) \end{aligned}$ In conclusion, particle's position at $t=4$ is $(-1018,-706.25)$.